A 30-06 caliber hunting rifle fires a bullet of
mass 0.0118 kg with a velocity of 505 m/s to
the right. The rifle has a mass of 6.15 kg.
What is the recoil speed of the rifle as the
bullet leaves the rifle?
Answer in units of m/s.
MV = Mv
(.0118)(505) = (6.15) (v)
(5.959) = 6.15v
v = .968943
V=mv/M. where V=?, m=0.0118kg, v=505m/s, M=6.15kg V=0.0118(505)/6.15 V=5.959/6.15 V=0.969m/s
Well, that's quite an explosive situation you've got there! Let's calculate the recoil speed of the rifle as the bullet leaves it.
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. So, when the bullet is fired to the right with a certain velocity, the rifle experiences a recoil effect to the left.
To find the recoil speed, we can use the conservation of momentum. The total momentum before the bullet is fired is zero since the rifle is initially at rest. After the bullet is fired, the total momentum remains zero because the momentum of the bullet to the right is equal and opposite to the momentum of the rifle to the left.
We can set up the equation:
(mass of bullet x velocity of bullet) + (mass of rifle x velocity of rifle) = 0
(0.0118 kg x 505 m/s) + (6.15 kg x velocity of rifle) = 0
Now, let's solve for the recoil velocity of the rifle:
(0.0118 kg x 505 m/s) = -(6.15 kg x velocity of rifle)
velocity of rifle = -(0.0118 kg x 505 m/s) / (6.15 kg)
velocity of rifle = -0.096 m/s
So, the recoil speed of the rifle as the bullet leaves it is approximately 0.096 m/s to the left.
I hope this answer didn't recoil you too much!
To find the recoil speed of the rifle, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after an event remains constant.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v). Mathematically, momentum is given by:
p = m * v
Before the bullet leaves the rifle, the initial momentum of the system (rifle + bullet) is zero since they are at rest. Therefore, the total momentum after the bullet leaves the rifle should also be zero.
Let's assume the recoil speed of the rifle is v_r (to the left) and the speed of the bullet is v_b (to the right).
The initial momentum (before the bullet leaves) is:
m_rifle * 0 + m_bullet * 0 = 0
The final momentum (after the bullet leaves) is:
m_rifle * v_r + m_bullet * v_b = 0
Substituting the given values:
6.15 kg * v_r + 0.0118 kg * (-505 m/s) = 0
Now, let's solve for v_r (recoil speed of the rifle):
6.15 kg * v_r = 0.0118 kg * (505 m/s)
v_r = (0.0118 kg * 505 m/s) / 6.15 kg
v_r ≈ 0.096 m/s
Therefore, the recoil speed of the rifle as the bullet leaves is approximately 0.096 m/s.
To find the recoil speed of the rifle, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, assuming no external forces are acting on the system.
In this case, we have a rifle and a bullet. The bullet is fired to the right with a certain velocity, and we want to find the recoil speed of the rifle.
Let's denote the initial velocity of the rifle as v_rifle_initial, the final velocity of the rifle as v_rifle_final, the mass of the bullet as m_bullet, and the velocity of the bullet as v_bullet.
The principle of conservation of momentum can be expressed as:
(m_rifle * v_rifle_initial) + (m_bullet * v_bullet) = (m_rifle * v_rifle_final)
Where:
m_rifle is the mass of the rifle (6.15 kg)
m_bullet is the mass of the bullet (0.0118 kg)
v_rifle_initial is the initial velocity of the rifle (unknown)
v_rifle_final is the final velocity of the rifle (unknown)
v_bullet is the velocity of the bullet (505 m/s)
We need to solve for v_rifle_final.
Since the bullet is fired to the right (positive direction), we can assign a positive sign to v_bullet. Similarly, the recoil velocity of the rifle will be in the opposite direction (to the left), so we assign a negative sign to v_rifle_final.
The equation becomes:
(6.15 kg * v_rifle_initial) + (0.0118 kg * 505 m/s) = (6.15 kg * -v_rifle_final)
Now we can solve for v_rifle_final:
6.15 kg * v_rifle_initial + 0.0118 kg * 505 m/s = -6.15 kg * v_rifle_final
Rearranging the equation:
6.15 kg * (v_rifle_initial + v_rifle_final) = -0.0118 kg * 505 m/s
Dividing both sides by 6.15 kg:
v_rifle_initial + v_rifle_final = (-0.0118 kg * 505 m/s) / 6.15 kg
Simplifying:
v_rifle_initial + v_rifle_final = -0.096 m/s
Now, since the bullet is being fired and the rifle is recoiling in the opposite direction, we can assume that the initial velocity of the rifle (v_rifle_initial) is 0.
Thus, the equation becomes:
0 + v_rifle_final = -0.096 m/s
Simplifying:
v_rifle_final = -0.096 m/s
So, the recoil speed of the rifle as the bullet leaves the rifle is approximately 0.096 m/s to the left.